I was listless today and not very productive. I made multiple stabs at sitting down to do exercises, but I couldn’t concentrate on them. In the end, I watched some videos and started writing up my proof of the inscribed triangle proposition from last week. I hoped to post that this evening, but the writing took longer than I was able to spend blogging today. I’m trying to make this proof more readable than my other recent geometric proofs, and that absorbs a lot of time. Let me know if you think such an effort is worthwhile. It’s felt to me as if the recent proofs tended toward being obscure jumbles of letters.
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My back is somewhat better today. I spent an hour on exercises in Analysis with an Introduction to Proof and I plan to read some more of Infinite Powers this evening, as well.
I’m also ready to prove the proposition about inscribed angles from my post Groundwork. It turns out that proposition falls right out of another that reader Tim McL suggested starting with when I said I knew little about the geometry of circles.
Proposition: If a central angle and an inscribed angle are subtended by the same arc of a circle, then the measure of the central angle is twice that of the inscribed angle.
Let $A$ and $B$ be points on a circle with center $D$.
Draw the central angle subtended by arc $AB$.
Draw an inscribed angle also subtended by arc $AB$ and let its vertex be called $C$.
Join $\overline {CD}$.
Now suppose that the sides of the two angles intersect only at $A$ and $B$ (as in the figure above).
Since $\overline{BD}$ and $\overline{CD}$ are both radii of the circle, $BD=CD$, from which it follows that $\angle CBD = \angle BCD$ (as I’ve proven previously).
Similarly, $\overline{AD}$ and $\overline{CD}$ are both radii of the circle, so $AD=CD$ and $\angle CAD = \angle ACD$.
Futhermore, since the sum of the angles of a triangle is $\pi$ radians, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.
It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.1
Combining these equations will yield $\angle ADB = 2(\angle BCD + \angle ACD)$.
But $\angle ACB = \angle BCD + \angle ACD$, so $\angle ADB = 2\angle ACB$, and the theorem is proven for this case.
Suppose instead that sides of the angles intersect other than at $A$ and $B$, as shown above. (Note that this argument does apply in the edge case where $\overline{BC}$ intersects $\overline{AD}$ at $D$.)
Then, $\angle ACB =\angle ACD-\angle BCD$ and $\angle ADB = \angle BDC – \angle ADC$.
As above, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.
Combining these equations will yield $\angle ADB = 2(\angle ACD – \angle BCD)$.
Hence $\angle ADB = 2\angle ACB$, and the theorem is proven for this case, as well.
After proving this proposition, I looked up how Euclid did it. His solution was largely the same. It was slightly more elegant, using the property of triangles illustrated below to cut out some of the algebra, but not so elegant as to dispense with the cases.
Proposition: If two inscribed angles are subtended by the same arc of a circle, then they are equal.
I’m just going to sketch a proof of this. By the proposition above, the blue and green inscribed angles subtended by arc $AB$ both have half the measure of the red central angle subtended by the same arc. Therefore, they must be equal to each other.
Footnotes
- Do I need to prove this? I have an idea of how I would do it, but it seem so fundamental. ↩︎
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My back pain continued today. I did manage to get some things done, but was too tired to do any math besides reading more of Infinite Powers.
One thing Strogatz relates in that book is that Greek mathematicians emphasized geometry partly because they had no way of expressing irrational numbers, such as $\pi$, except in geometric terms: as lengths, areas, and their ratios. If you think about it, that has not changed much. We may have discovered more contexts in which $\pi$ or $\phi$ crop up, but we can still only express them exactly as solutions to particular problems. Any effort to write them down in isolation is only an approximation.
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I am having an issue with my Internet connection today, so I’m afraid I will have to put off sharing my proof of the proposition about two inscribed angels subtended by the same arc.
I’m also having an issue with my back that is proving very draining. After doing other things that needed to be done, I only had the energy left to watch some math videos, not do more analysis exercises. Hopefully tomorrow will be better. (Said Olly, again.)
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Today I worked on exercises in Analysis with an Introduction to Proof and did a little mathematical puttering.
I’m also ready today to start sharing my work on the proposition about the inscribed equilateral triangle. I’ll start by proving that a triangle is equilateral if and only if its three angles are equal.
First, recall from my post Revenge of the Squares, Part 1 that the angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal. The proof of the following related lemma is very similar.
Lemma: If two angles of a triangle are equal, then the sides that subtend them are also equal.
Consider the triangle $\triangle ABC$ with $\angle ACB\cong\angle ABC$.
Draw a line bisecting $\angle BAC$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.
Now, by the angle-angle-side property, $\triangle ACD\cong\triangle ABD$, since $\angle ACB\cong\angle ABC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.
Therefore, $AC=AB$. $\overline {AC}$ and $\overline {AB}$ are the sides that subtend the equal angles $\angle ACB$ and $\angle ABC$, so the lemma is proven.
(Note that I have added AAS to my list of triangle congruence properties to prove.1)
Proposition: A triangle is equilateral if and only if its three angles are equal.
Consider a triangle $\triangle ABC$.
Assume that $\triangle ABC$ is equilateral.
Then, since $AB=AC$, $\angle ACB\cong\angle ABC$, by the theorem cited above. Similarly, since $AC=BC$, $\angle ABC\cong\angle BAC$.
It follows that $\angle ACB\cong\angle ABC\cong\angle BAC$.
Thus, if $\triangle ABC$ is equilateral, then its three angles are equal.
Assume instead that the three angles of $\triangle ABC$ are equal.
Then, by the lemma, since $\angle ACB\cong\angle ABC$, $AB=AC$. Similarly, since $\angle ABC\cong\angle BAC$, $AC=BC$.
It follows that $AB=AC=BC$.
Thus, if the three angles of $\triangle ABC$ are equal, then it is equilateral.
This proves the equivalence.
Stay tuned tomorrow for some circle geometry.
- To do. ↩︎
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My mental state remained good today, but I also slept a lot. I did some housekeeping on the blog to make it easier for me to return to questions I’ve raised but not answered. My study time was devoted to work on the proposition about inscribed angles subtended by the same arc and to considering whether and how my other recent proofs depend on the parallel postulate. The proof of the inscribed triangle proposition does depend on it, I believe. It is not necessary to the proof that the sum of the angles of a triangle is $\pi$ radians, but it is necessary that it is the same for all triangles, which is not the case in some systems of geometry.
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Today was devoted to fun with geometry (as was some time late last night). I came up with a simpler proof of the proposition about the inscribed equilateral triangle, which I’m pleased about; the first one was a bit gnarly. I also worked on the lemmas to accompany it. It needs most of the same ones, but not the one about parallelograms.
Here are before and after shots showing the working figure for my first proof and that for my second. One proof required four new lines and two new named points. The other involved just one one of each.
(I’m looking forward to finishing this notebook so that I can move to an unruled one.)
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I spent quite a bit of time today working on the inscribed equilateral triangle problem. I experimented both by hand and using the Desmos geometry tool, which allows you to vary parameters in your drawings. At this point, I’ve come up with a proof that convinces me. There is one step that needs to be formalized and a couple that depend on facts that I know to be true but would like to prove. These include that the sides of a parallelogram are pairwise equal1 and that a triangle is equilateral if and only if all of its angles are equal. The argument also depends heavily on a fact I just learned today, conjecturing it based on experimentation and then confirming it with a Google search: inscribed angles subtended by the same arc of a circle are equal (as shown below). I’m not sure whether I will be able to prove that or not. My education in geometry focused almost exclusively on lines and triangles, and I know little of circle geometry.
(As readers may have guessed, I’m feeling much better today.)
- To do. ↩︎
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Today I worked for as long as I could on the proposition from yesterday about the inscribed equilateral triangle. I’m feeling a little better, but my mind is still sluggish, so I didn’t make much progress.
I’m wondering whether proving the proposition for a special case first would be helpful. The edge cases where $D$ is the same point as $A$ or $C$ are trivial. The case where $AD = CD$ might provide some insight. On the other hand, using any of the unique characteristics of that case to prove it could leave me as far from a general proof as before.
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Things continued difficult today, but I felt a bit better after dinner. I watched four videos from the Numberphile channel, notably one about proving a geometric relationship called Ptolemy’s Theorem by a technique called inversion of the plane. It was a great video, but quite involved. I would not recommend it to those without a strong interest in math.
When I am feeling better, I would like to have a go at proving Ptolemy’s Theorem by another method.1 The lecturer in the video said that it can be done using plane geometry, provided you are clever, and can also be done by crunching trig ratios, something I have some experience in. I’d also like to prove, by some other method, a fact that can be proven using Ptolemy’s Theorem: that in the figure shown below, the distance from $D$ to $B$ is the sum of the distances from $D$ to $A$ and $C$.
A cool thing about Ptolemy’s Theorem is that the Pythagorean Theorem falls right out of it. In fact, you could say that the Pythagorean Theorem is a special case of Ptolemy’s. This means that Ptolemy’s Theorem must only be true if you assume the Parallel Postulate. I wonder if that is the case for the proposition above.
- To do. ↩︎